Define the point on the x-axis.

A point on the x-axis has a y-coordinate of 0. Let the required point be $P(x, 0)$.
The given points are $A(2, -5)$ and $B(-2, 9)$.

The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $D = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
Since point $P$ is equidistant from $A$ and $B$, we have $PA = PB$.
Squaring both sides, we get $PA^2 = PB^2$.

Subtract $x^2$ from both sides:

Subtract $4x$ from both sides:

Subtract 29 from both sides:

Divide by $-8$:

The point on the x-axis is $P(x, 0)$, so substituting $x = -7$, we get the point as $(-7, 0)$.

The point on the x-axis equidistant from $A(2, -5)$ and $B(-2, 9)$ is $\boxed{(-7, 0)}$.